Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

G=2.7, YW=9.81 ,
W=22/100=0.22.

Bulk unit wight=19.62
1. Yd=19.62/(1+w)=19.62/(1+0.22)=16.08
2. void ratio (e)=(G*Yw/Yd)-1=(2.7*9.81/16.08)-1=0.647
3. saturatted unit weight (s)=G*W/e=2.7*0.22/0.647=0.918
0.918*100=91.8%.

Yd = (1+GYw)÷(1+wG÷Sr).

Thanks @Gilbert Khakhlari.

Very well explained, Thanks @Ujjval.

W=.1, bulk density = 2g/cc.
Dry density = 2/1 +.1 = 1.8g/cc.
e = [(2.7 * .1)/1.8-1] = .485.
S = (2.7 * 0.1)/.485 = .556.
OR
S = (.556*100)%.

B is the answer.

Given That,

Water Content w = 0.1
Unit Wt of Soil = 2.0 g/cc.
Sp. Gravity of Soil = 2.7.
Unit Wt of water = 1 g/cc.

Dry Unit Wt of Soil = Bulk unit wt/ (1+ w).
= 2/(1+0.1).
=1.818 g/cc.


Now,

Void Ratio (e)=(G * Unit Wt of Water/dry unit wt ) - 1.
= (2.7 * 1 / 1.818) - 1.
= 0.4851 .
Now,

Degree of Saturation (S) = G*w/e.
=2.7*0.1/0.4851.
=0.556.
=55.6% Ans.

But the answer showing is 69.6%.

Partially saturated soil samples obtained from and Earth fill as a natural moisture contain 22%of Unit Weight of 19.62 kn/m^3. Given a specific gravity 2.7 and unit of water 9.81kn/m^3..determine the degree of saturation, voids ratio, saturated unit weight, dry unit weight of soil.

Can anyone please solve this?

Yeah, I also got 55.6.

Great explanation @Sayandip Basak.